IIT JEE 1974 Mathematics Question Paper with Answer and Solution

(B) Let $S_n = 6 + 66 + 666 + \dots$ up to $n$ terms.
$S_n = 6(1 + 11 + 111 + \dots + n \text{ terms})$
$S_n = \frac{6}{9}(9 + 99 + 999 + \dots + n \text{ terms})$
$S_n = \frac{2}{3}((10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1))$
$S_n = \frac{2}{3}((10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times}))$
Using the sum of a geometric progression formula $\frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{10 - 1} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10(10^n - 1)}{9} - n \right)$
$S_n = \frac{2}{3} \left( \frac{10^{n+1} - 10 - 9n}{9} \right)$
$S_n = \frac{2(10^{n+1} - 9n - 10)}{27}$.

(D) Let the four numbers be $\frac{a}{r}, a, ar, x$.
Since the last three numbers $a, ar, x$ are in $A.P.$ with common difference $d = 6$,we have $ar - a = 6$ and $x - ar = 6$.
Thus,$x = ar + 6$.
Since $ar = a + 6$,we have $x = (a + 6) + 6 = a + 12$.
Given that the first and last numbers are equal,$\frac{a}{r} = x = a + 12$.
From $ar - a = 6$,we get $a(r - 1) = 6$,so $r - 1 = \frac{6}{a}$,which means $r = 1 + \frac{6}{a} = \frac{a + 6}{a}$.
Substituting $r$ into $\frac{a}{r} = a + 12$:
$\frac{a}{(a + 6)/a} = a + 12$
$\frac{a^2}{a + 6} = a + 12$
$a^2 = (a + 12)(a + 6)$
$a^2 = a^2 + 6a + 12a + 72$
$0 = 18a + 72$
$18a = -72$
$a = -4$.
The first number is $\frac{a}{r} = a + 12 = -4 + 12 = 8$.

(C) Let the common root be $y$.
Then,$y^2 + py + q = 0$ and $y^2 + \alpha y + \beta = 0$.
Subtracting the two equations:
$(y^2 + py + q) - (y^2 + \alpha y + \beta) = 0$
$y(p - \alpha) + (q - \beta) = 0$
$y(p - \alpha) = \beta - q$
$y = \frac{\beta - q}{p - \alpha} = \frac{q - \beta}{\alpha - p}$.
Alternatively,using the method of cross-multiplication for the system:
$\frac{y^2}{p\beta - q\alpha} = \frac{y}{q - \beta} = \frac{1}{\alpha - p}$.
From the second and third terms,$y = \frac{q - \beta}{\alpha - p}$.
From the first and second terms,$y = \frac{p\beta - q\alpha}{q - \beta}$.
Thus,the common root is $\frac{q - \beta}{\alpha - p}$ or $\frac{p\beta - \alpha q}{q - \beta}$.

(D) Let $f(x) = x^4 - px^2 + q$. Since $x^2 - 3x + 2$ is a factor of $f(x)$,the roots of $x^2 - 3x + 2 = 0$ must also be roots of $f(x) = 0$.
$x^2 - 3x + 2 = (x - 1)(x - 2) = 0$,so the roots are $x = 1$ and $x = 2$.
For $x = 1$: $1^4 - p(1)^2 + q = 0$ $\Rightarrow 1 - p + q = 0$ $\Rightarrow p - q = 1$ (Equation $i$).
For $x = 2$: $2^4 - p(2)^2 + q = 0$ $\Rightarrow 16 - 4p + q = 0$ $\Rightarrow 4p - q = 16$ (Equation $ii$).
Subtracting $(i)$ from (ii): $(4p - q) - (p - q) = 16 - 1$ $\Rightarrow 3p = 15$ $\Rightarrow p = 5$.
Substituting $p = 5$ into $(i)$: $5 - q = 1 \Rightarrow q = 4$.
Thus,$(p, q) = (5, 4)$.

(B) To form a group,we must select a subset of the available balls.
For the $5$ different green balls,the number of ways to select at least one green ball is $2^5 - 1 = 31$.
For the $4$ different blue balls,the number of ways to select at least one blue ball is $2^4 - 1 = 15$.
For the $3$ different red balls,we can either select or not select each ball,which gives $2^3 = 8$ ways.
Since the selections are independent,the total number of ways to form the group is $31 \times 15 \times 8 = 3720$.

(D) Given expression: $\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$
$= \frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{2(\frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
Using $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
$= \frac{2(\sin 30^\circ \cos 10^\circ - \cos 30^\circ \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{2 \sin(30^\circ - 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
$= \frac{2 \sin 20^\circ}{\sin 10^\circ \cos 10^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{4 \sin 20^\circ}{2 \sin 10^\circ \cos 10^\circ}$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{4 \sin 20^\circ}{\sin 20^\circ} = 4$.

(C) We use the identity $\tan \theta \tan(60^\circ - \theta) \tan(60^\circ + \theta) = \tan 3\theta$.
Given expression is $E = \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ$.
Rearranging the terms,we have $E = \tan 60^\circ (\tan 20^\circ \tan 40^\circ \tan 80^\circ)$.
Note that $\tan 80^\circ = \tan(60^\circ + 20^\circ)$ and $\tan 40^\circ = \tan(60^\circ - 20^\circ)$.
Using the identity for $\theta = 20^\circ$:
$\tan 20^\circ \tan(60^\circ - 20^\circ) \tan(60^\circ + 20^\circ) = \tan(3 \times 20^\circ) = \tan 60^\circ$.
Substituting this back into the expression for $E$:
$E = \tan 60^\circ \times \tan 60^\circ = \sqrt{3} \times \sqrt{3} = 3$.

(B) Let the vertices be $A(0, 0)$,$B(2, -1)$,and $C(1, 3)$.
$1$. Find the equation of altitude from $A$ to $BC$:
Slope of $BC = \frac{3 - (-1)}{1 - 2} = \frac{4}{-1} = -4$.
The slope of the altitude $AD$ is the negative reciprocal of the slope of $BC$,which is $\frac{1}{4}$.
The equation of $AD$ passing through $A(0, 0)$ is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $x - 4y = 0$ ..... $(i)$.
$2$. Find the equation of altitude from $B$ to $AC$:
Slope of $AC = \frac{3 - 0}{1 - 0} = 3$.
The slope of the altitude $BE$ is the negative reciprocal of the slope of $AC$,which is $-\frac{1}{3}$.
The equation of $BE$ passing through $B(2, -1)$ is $y - (-1) = -\frac{1}{3}(x - 2)$,which simplifies to $3(y + 1) = -(x - 2)$ $\Rightarrow 3y + 3 = -x + 2$ $\Rightarrow x + 3y + 1 = 0$ ..... $(ii)$.
$3$. Solve the system of equations $(i)$ and $(ii)$:
From $(i)$,$x = 4y$. Substitute this into $(ii)$:
$4y + 3y + 1 = 0$ $\Rightarrow 7y = -1$ $\Rightarrow y = -\frac{1}{7}$.
Then $x = 4(-\frac{1}{7}) = -\frac{4}{7}$.
Thus,the orthocentre is $\left( -\frac{4}{7}, -\frac{1}{7} \right)$.

(A) The equation of any line passing through $(3, -2)$ is $y + 2 = m(x - 3)$ ..... $(i)$
The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.
The angle $\theta = 60^o$ between the lines is given by $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
Substituting the values,$\tan 60^o = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \right| \implies \sqrt{3} = \left| \frac{m + \sqrt{3}}{1 - m\sqrt{3}} \right|$.
Case $1$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = \sqrt{3} \implies m + \sqrt{3} = \sqrt{3} - 3m \implies 4m = 0 \implies m = 0$.
Case $2$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = -\sqrt{3} \implies m + \sqrt{3} = -\sqrt{3} + 3m \implies 2m = 2\sqrt{3} \implies m = \sqrt{3}$.
Substituting $m = 0$ in $(i)$: $y + 2 = 0(x - 3) \implies y + 2 = 0$.
Substituting $m = \sqrt{3}$ in $(i)$: $y + 2 = \sqrt{3}(x - 3) \implies y + 2 = \sqrt{3}x - 3\sqrt{3} \implies \sqrt{3}x - y - 2 - 3\sqrt{3} = 0$.

(A) We have $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}}$.
Using the expansion $\tan x = x + \frac{x^3}{3} + \dots$ and $\sin x = x - \frac{x^3}{6} + \dots$,we get:
$\mathop {\lim }\limits_{x \to 0} \frac{{(x + \frac{x^3}{3} + \dots) - (x - \frac{x^3}{6} + \dots)}}{{{x^3}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x^3}{3} + \frac{x^3}{6}}}{{{x^3}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{3x^3}{6}}}{{{x^3}}} = \frac{1}{2}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes where the sum of the points is $7$ are: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
The number of favourable outcomes is $6$.
Therefore,the probability is $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36}$.

(B) To evaluate $I = \int x^2 \sin 2x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$.
Let $u = x^2$ and $dv = \sin 2x \, dx$.
Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$.
Applying the formula:
$I = x^2 \left(-\frac{\cos 2x}{2}\right) - \int \left(-\frac{\cos 2x}{2}\right) (2x \, dx)$
$I = -\frac{x^2 \cos 2x}{2} + \int x \cos 2x \, dx$.
Now,apply integration by parts again for $\int x \cos 2x \, dx$:
Let $u = x$ and $dv = \cos 2x \, dx$.
Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\int x \cos 2x \, dx = x \left(\frac{\sin 2x}{2}\right) - \int \frac{\sin 2x}{2} \, dx$
$= \frac{x \sin 2x}{2} - \frac{1}{2} \left(-\frac{\cos 2x}{2}\right) = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}$.
Substituting this back into the expression for $I$:
$I = -\frac{x^2 \cos 2x}{2} + \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + c$.